Question: Is ${24933}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {24933}= &&{2}\cdot10000+ \\&&{4}\cdot1000+ \\&&{9}\cdot100+ \\&&{3}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {24933}= &&{2}(9999+1)+ \\&&{4}(999+1)+ \\&&{9}(99+1)+ \\&&{3}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {24933}= &&\gray{2\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {2}+{4}+{9}+{3}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${24933}$ is divisible by $3$ if ${ 2}+{4}+{9}+{3}+{3}$ is divisible by $3$ Add the digits of ${24933}$ $ {2}+{4}+{9}+{3}+{3} = {21} $ If ${21}$ is divisible by $3$ , then ${24933}$ must also be divisible by $3$ ${21}$ is divisible by $3$, therefore ${24933}$ must also be divisible by $3$.